Evaluation of statements:

• Doping pure silicon with trivalent impurities (like Boron, Aluminium, Indium) creates excess holes, forming a p-type semiconductor (Correct statement).
• In an n-type semiconductor, doping with pentavalent impurities introduces free conduction electrons as majority carriers, whereas holes are minority carriers. Therefore, the statement claiming majority carriers are holes is incorrect (NOT correct statement).
• In a p-type semiconductor, holes are majority carriers and electrons are minority carriers (Correct statement).
• In intrinsic semiconductors, higher temperature breaks covalent bonds to generate electron-hole pairs, increasing conductivity and thus decreasing resistivity (Correct statement).

Circuit Analysis using Boolean Algebra:

• The top NAND gate has both of its inputs tied together to input $A$. Its output is:
  $$Y_1 = \overline{A \cdot A} = \bar{A}$$
• The bottom NAND gate has both of its inputs tied together to input $B$. Its output is:
  $$Y_2 = \overline{B \cdot B} = \bar{B}$$
• These two inverted outputs are fed into the third NAND gate. The final output $Y$ is:
  $$Y = \overline{Y_1 \cdot Y_2} = \overline{\bar{A} \cdot \bar{B}}$$
• Applying De Morgan's Law ($\overline{X \cdot Y} = \bar{X} + \bar{Y}$):
  $$Y = \overline{\bar{A}} + \overline{\bar{B}} = A + B$$
  This is precisely the Boolean expression for an OR gate.

Circuit Analysis:

• Let node $B$ be at reference potential $V_B = 0\,\text{V}$.
• The battery maintains the top wire at $24\,\text{V}$.
• The junction between the two vertical $8\,\text{k}\Omega$ resistors is at a positive potential relative to $B$.
• Because the p-side (anode) of the diode is connected to this positive junction and the n-side (cathode) connects to node $A$ (which is connected to $0\,\text{V}$ via the right $8\,\text{k}\Omega$ resistor), the diode is FORWARD BIASED.
• Assuming an ideal diode, it acts as a short circuit with zero resistance.
  6. Consequently, the right branch $8\,\text{k}\Omega$ resistor is connected in parallel with the bottom $8\,\text{k}\Omega$ resistor across terminals $A$ and $B$:
  $$R_{\text{parallel}} = \frac{8\,\text{k}\Omega \times 8\,\text{k}\Omega}{8\,\text{k}\Omega + 8\,\text{k}\Omega} = 4\,\text{k}\Omega$$
  7. The total equivalent circuit resistance is:
  $$R_{\text{total}} = 8\,\text{k}\Omega + R_{\text{parallel}} = 8 + 4 = 12\,\text{k}\Omega$$
  8. Using the voltage divider rule, the potential across the parallel combination (between $A$ and $B$) is:
  $$V_{AB} = 24\,\text{V} \times \frac{4\,\text{k}\Omega}{12\,\text{k}\Omega} = 24 \times \frac{1}{3} = 8\,\text{V}$$